# L'Hôpital's Rule

## Contents

## Discovered By

Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli.

## The Rule

For or case, the limit where and are the first derivatives of and , respectively.

## Examples

## Proof of l'HÃ´pital's rule

A standard proof of l'HÃ´pital's rule uses Cauchy's mean value theorem. l'HÃ´pital's rule has many variations depending on whether *c* and *L* are finite or infinite, whether *f* and *g* converge to zero or infinity, and whether the limits are one-sided or two-sided.

### Zero over zero

Suppose that *c* and *L* are finite and *f* and *g* converge to zero.

First, define (or redefine) *f*(*c*) = 0 and *g*(*c*) = 0. This makes *f* and *g* continuous at *c*, but does not change the limit (since, by definition, the limit does not depend on the value at the point *c*).

Since exists, there is an interval (*c* − *Î´*, *c* + *Î´*) such that for all *x* in the interval, with the possible exception of *x* = *c*, both and exist and is not zero.

If *x* is in the interval (*c*, *c* + *Î´*), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [*c*, *x*] (and a similar statement holds for *x* in the interval (*c* − *Î´*, *c*)). The mean value theorem implies that *g*(*x*) is not zero (since otherwise there would be a *y* in the interval (*c*, *x*) with ). Cauchy's mean value theorem now implies that there is a point *Î¾*_{x} in (*c*, *x*) such that

If *x* approaches *c*, then *Î¾*_{x} approaches *c* (by the squeeze theorem ). Since exists, it follows that

### Infinity over infinity

Suppose that *L* is finite, *c* is positive infinity, and *f* and *g* converge to positive infinity.

For every *Îµ* > 0, there is an *m* such that

The mean value theorem implies that if *x* > *m*, then *g*(*x*) â‰ *g*(*m*) (since otherwise there would be a *y* in the interval (*m*, *x*) with ). Cauchy's mean value theorem applied to the interval [*m*, *x*] now implies that

Since *f* converges to positive infinity, if *x* is large enough, then *f*(*x*) â‰ *f*(*m*). Write

Now,

- $\begin{align}

& \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align} allowed only in paragraph mode.)

For *x* sufficiently large, this is less than *Îµ* and therefore

- *

- Note: Steps are missing.